CODEWITHSUNDEEP
htmlhaml
Sean Ahrens
Sean Ahrens

Reputation: 1025

How can I conditionally wrap some HAML content in a tag?

How can I use a condition to decide whether to output a surrounding tag in HAML? I'm trying to create the DRY version of the code below.

- if i_should_link 
  %a{:href => url}
   .foo
     .block
       .of  
         .code
- else
  .foo
    .block
      .of  
        .code

Upvotes: 39

Views: 11600

Answers (2)

Nathan Long
Nathan Long

Reputation: 125892

I think odin's suggestion to use a partial is probably the best in most situations.

However, as an alternate solution, I found a thread where Nathan Weizenbaum suggested defining this method:

def haml_tag_if(condition, *args, &block)
  if condition
    haml_tag *args, &block
   else
     yield
   end
end

Whatever is in the block would always be rendered, but the wrapping tag would appear or not based on the condition.

You would use it as follows:

- haml_tag_if(planning_to_mail?, :div, :id => 'envelope') do
   %p I'm a letter

If planning_to_mail? evaluates true, you'd get:

<div id="envelope">
  <p>I'm a letter</p>
</div>

If it evaluates false, you'd get:

<p>I'm a letter</p>

He floated the idea of adding this to Haml::Helpers, but that doesn't appear to have happened yet.

Upvotes: 27

twe4ked
twe4ked

Reputation: 2892

You could use a partial.

foo.html.haml

- if i_should_link
  %a{:href => url}
    = render 'bar'
- else
  = render 'bar'

_bar.html.haml

.foo
  .block
    .of
      .code

Edit: Or you could use content for, I guess this is better because it keeps it all in the same file.

- if i_should_link
  %a{:href => url}
    = yield :foobar
- else
  = yield :foobar

- content_for :foobar do
  .foo
    .block
      .of
        .code

Upvotes: 37

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